[Author Prev][Author Next][Thread Prev][Thread Next][Author Index][Thread Index]

Re: Correcting for 80% VE - just more questions then

QSHIPQ@aol.com wrote:

> Based on a couple of recent posts, .  The givens are as follows:
>  S.O.c claims and audi facts, THESE STAY THE SAME @:
>  *  26psi in 2nd thru 5th
>  *  2.5 bar PT (given)
>  *  16+psi in 1st gear
>  *  .5 PD accross modded IC
>  *  Audi --> RS2 gets 310hp @ 460CFM (given)
>  *  1325 EGT
>  *  Audi --> MC motor = 134ci (given)
>  *  Stock WG spring
>  *  "...An RSR turbo, a Ned Ritchie turbo, blows the RS2 into the
> weeds"
>  *  "... the car now will outrun just about any other MC engine
> around.
> Correcting my post to the "homework" section, with a 80% VE the
> questions
> become:
>  The homework would be
>  A) How does a 136ci motor flow 482CFM <was 513> in 10v trim?  Can it
> stock?
>  B)  Why such a large PR?  I thought Density Ratio was more the goal?
>  C)  Given a 2.79 PR vs a 2.50PR with the same output, which turbo
> should one
> use?
>  D)  Then, how does an S.O.c. car "blow the RS2 wannabes (me by
> definition I
> suppose) into the weeds?"
>  E)  How does a PT that can control boost to 2.50PR by definition,
> control
> boost to 2.79PR with a stock WG spring?
> Another piece of math from any turbocharger book shows:
> Baseline CFM * Pressure Ratio = boosted engine CFM  (look hard so far,
> everyone, there are NO mods here)
> So:
> 173 * 2.79 = 482CFM
> Then a10vt motor with a 2.79PR flows the corrected 482CFM.  Some more
> calculations from someone here can determine  what the runner and
> effective
> valve size have to be (this is where I get paid, so someone else can
> do it).
>  Can it be done stock?  IF the runner and effective valve size HAS to
> be
> increased over stock, then 482CFM is now min, cuz changing runner
> and/or
> effective valve size from a stock motor, increases it's baseline
> efficiency>larger CFM baseline AND boosted.  Cool eh?  So really,
> Motor Mods
> aren't going to help in the argument at all.

> Interesting questions Bruce.  What you are missing is a couple of
> things that
> might/should concern you.  2.79PR =26psi gives you a BMEP (someone
> else can
> calculate that, I think it's too funny to put a number to) on a 7.8CR
> motor,
> what does that mean for you with 8.4?  Now, the accepted rule of thumb
> is:
> Turbocharged cars perform best with the lowest Pressure Ratio at the
> Highest
> Density ratio (this assumes CFM to be a constant)
> (To confirm this, see mathmatical example in IC efficiency section of
> my
> original post, it is correct as printed.)
> So, a hybrid k26/27 turbo needs 2.79PR (26psi corrected) to put out
> the same
> flow as a RS2 at 2.50PR (21.75 corrected), which should one choose
> given the
> above ROT?  Specifically, how does a 2.79PR turbo "blow the RS2 into
> the
> weeds, when, in fact, the PR is higher, and the DR is lower too, by
> definition.
> What does this mean to the confused?  It really doesn't change my
> original
> conclusion that the wrong turbo is doing the wrong job on the wrong
> engine.
>  A box stock RS2, in this case, MODS ASIDE, will perform on Eric's car
> better
> than the one he has.  By definition:  Turbocharger Efficiency is
> higher, BMEP
> is lower, PR is lower, and Density Ratio is higher.  These are all
> win/win
> concepts to a turbocharged application from what I know, and most
> gurus will
> tell you.

1. Why?  Are you saying that a larger turbo is less efficient at
producing boost than a smaller turbo?  Correct me if I am wrong, but I
thought that larger turbos were more efficient at making large
quantities of boost and smaller turbos spool up faster at the loss of
top end boost.  One might make the assumption that if a smaller turbo is
better then smallest is better yet.  I know that is not true.  SO why
would a smaller RS2 make more dense air than a larger more efficient at
generating high boost turbo?

> WAS:
> "Baseline N/A motor airflow for a 134ci (MC) engine at 85% VE = 184CFM
> ********  CORRECTED TO:***************************
>  Baseline N/A motor airflow for a 136ci (MC) motor at 80% VE = 173CFM
> (Since 173 * 1.43PR (stock MC turbo motor - bentley) = 244CFM
> boosted/1.5
> (est correction to HP) = 164.9 pretty good, thanks Orin.
> So, a claimed 2.79PR, by math means:
> 173 * 2.79 = 482CFM @ 5500 (RSR turbo 26/27 hybrid @ ?? efficiency @
> 5500
> Compare say the RS2 turbo
> 184 (remember I keep 85% efficiency here for the 4v head) * 2.50 =
> 460CFM
> Ok, and the stage II mods (assuming 2.0 bar)
> 173 * 2.00 = 340CFM    ****  346/1.5 correction = 230hp for you HP
> dudes, a
> little high in my estimation, your IC efficiency is probably dragging
> that
> down some.

2.  Why compare a 20v motor with the RS2 to a 10v motor with the hybrid
173*2.5=432.5CFM.  By my calculations that is an 11% drop in CFM and HP.

> I wholy accept that the 1325 EGT is darn good.  And really don't argue
> with
> the fuel part for now, minor issue in the grand scheme o' things.
> What really creates a question, is how a claim of 2.79bar can be had
> with a
> stock spring on a 2.5bar PT computer box.  That leaves only 1 of 4
> answers:
>  A) 26psi is guage error (but I have the same calibrated aircraft
> guage that
> S.O.c. does, and it can predict weather fronts, it's that good, and if
> it
> were failing, the pressure would read low, not high), discount that. 
> 2)
>  There is external boost control, not the claim, read above, or 3) The
> PT
> voltage is divided by a % so the computer is interpreting a lower PR. 
> I
> discount this for a variety of reasons, the most obvious being, you
> don't
> need a 2.5PT to do that, a 2.0PT can do the same thing.  4)  There
> isn't
> 2.79PR, is the claim, read above.

Ask those who programmed the box.  The other car has seen over 80in of
boost out of one chip.  Therefore, the boost has been governed by the
chip.  I do not happen to be privy to such details of the program.

> By definition, it has to be one of the above.  Why?  Cuz you can't
> program
> WGFV control over the measure limit of the PT, Period.   We also know
> that a
> 2.79PR means that there is no overboost protection using the same
> logic
> above, which means above 2.50PR the PT goes to a constant voltage
> (actualy
> it's usually less than that, but BOD).  So what stops the motor from
> going to
> 3bar?  We now know by the claim, it's not the Baseline Spring Pressure
> in the
> WG.   So, what do we assume?  Eric?  Randall?  Anyone?  Mysterious,
> magic,
> logic, claims vs math and physics now comes to a head.  It's
>  Trick or treat?   Smell my feet.
> Questions still remain, numbers (revised by request) still don't add
> up.

3.  What I still do not understand is how you differentiate the turbos
from one another  in your equations.  Lets say I have the Joe Schmoe
brand turbo that can just barely blow 2.50PR does the efficiency of the
turbo hold equal to the RS2 or is the DR lower because of the air being
hotter due to la ess efficient turbo?  I think any assumptions you are
making about the density of the air to the engine is just that an
assumption.  I think the bigger turbo works less to make the boost and
therefore has a higher air density than the RS2.  

Randall C. Markarian

1990 V8 Quattro
1996 Merc E320

Saint Louis, Missouri